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(D) tetrahedron has $4$ vertices and $6$ edges. Let the current enter at vertex $1$ and leave at vertex $2$.There is a direct wire between $1$ and $2$

Vedclass · Accepted Answer

$\frac{r}{2}$

Vedclass · Answer

(D) tetrahedron has $4$ vertices and $6$ edges. Let the current enter at vertex $1$ and leave at vertex $2$.There is a direct wire between $1$ and $2$ with resistance $r$.There are two paths from $1$ to $2$ through the other two vertices ($3$ and $4$):Path $1$: $1 \rightarrow 3 \rightarrow 2$ (two resistors in series, total $2r$)Path $2$: $1 \rightarrow 4 \rightarrow 2$ (two resistors in series, total $2r$)These two paths are in parallel with each other, so their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{2r} + \frac{1}{2r} = \frac{2}{2r} = \frac{1}{r}$, which means $R_p = r$.Finally, this $R_p$ is in parallel with the direct wire between $1$ and $2$ (which also has resistance $r$).Therefore, the total equivalent resistance $R_{eq}$ is $\frac{1}{R_{eq}} = \frac{1}{R_p} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$.Thus, $R_{eq} = \frac{r}{2}$.

Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

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